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2 years ago

A box is at rest on a slope

Physics

1 answer:

matrenka [14]2 years ago

6 0

Answer:

170 N

Explanation:

The following data were obtained from the question:

Mass (m) = 20 Kg

Angle (θ) = 30°

Acceleration due to gravity (g) = 9.8 m/s²

Weight (W) =?

The perpendicular component of the weight can be obtained by using the following formula:

W = mg × Cosθ

W = 20 × 9.8 × Cos 30

W = 196 × 0.866

W = 169.7 ≈ 170 N

Therefore, perpendicular component of the weight is 170 N

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Two boats leave the same port at the same time, with boat A traveling north at 15 knots (nautical miles per hour) and boat B tra

Mrrafil [7]

Answer:

The chance in distance is 25 knots

Explanation:

The distance between the two particles is given by:

$s^2 = (x_A - x_B)^2+(y_A - y_B)^2$ (1)

Since A is traveling north and B is traveling east we can say that their displacement vector are perpendicular and therefore (1) transformed as:

$s^2 = x_B^2+y_A^2$ (2)

Taking the differential with respect to time:

$\displaystyle{2s\frac{ds}{dt}= 2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt}}$ (3)

where $\displaystyle{\frac{dx_B}{dt}}=v_B$ and $\displaystyle{\frac{dx_A}{dt}}=v_A$ are the respective given velocities of the boats. To find $s$ and $x_B$ we make use of the given position for A, $y_A=30$ , the Pythagoras theorem and the relation between distance and velocity for a movement with constant velocity.

$\displaystyle{y_A = v_A\cdot t\rightarrow t = \frac{y_A}{v_A}=\frac{30}{15}=2 h$

with this time, we know can now calculate the distance at which B is:

$\displaystyle{x_B = v_B\cdot t= 20 \cdot 2 = 40\ nmi$

and applying Pythagoras:

$\displaystyle{s = \sqrt{x_B^2+y_A^2}=\sqrt{30^2 + 40^2}=\sqrt{2500}=50}$

Now substituting all the values in (3) and solving for $\displaystyle{\frac{ds}{dt} }$ we get:

$\displaystyle{\frac{ds}{dt} = \frac{1}{2s}(2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt})}\\\displaystyle{\frac{ds}{dt} = 25 \ knots}$

4 0

3 years ago

What are two reasons why a home computer scanner requires

alexandr1967 [171]

Answer: C and D

Explanation: a p e x

4 0

2 years ago

Read 2 more answers

Student pushes a 50 N block across the floor for a distance of 15 m how much work was done to move the block

Talja [164]

Answer:

750 J

Explanation:

We have a student that pushes a 50N block across the floor for a distance of 15m. The question is asking how much work was done to move the block.

To solve this, we must know that we are looking for a certain thing called joules. And to get the answer, we must follow the formula of W = FS

F being the force and S being the distance.

W = FS

W = (50)(15)

W = 750

Therefore, 750 joules is our answer.

7 0

3 years ago

A force of 20 N acts toward the east. Another force of 27 N acts at the same point toward the west. What is the magnitude and di

Jet001 [13]

Since we are working in one dimension (left right or East West), we don't need to worry about angles! It's just simply a matter of adding things up!

First list out all the forces and add negative (-ive) signs to each of the 'west' forces like this.

20 East + (-27 West) + ? = 10 East

so it's easy to see that 20 + (-27) = -7
So to get to 10 from -7 just do the sum to get 17.
Since 17 is not negative it must be in the direction of East.
So the answer is:
Magnitude = 17 N
Direction = toward the East

8 0

3 years ago

Someone help please 1 You are cycling along a flat road. a Are you moving up or down? b What is the force that is pulling you

9966 [12]

Answer:

$\gamma \: 1a. \: i \: am \: moving \: downwards \: . \\ b. \: gravity \: is \: pulling \: me \: downwards \: \\ c. \: the \: force \: is \: acting \: downwards \: \\ the \: answer \: to \: the \: question \: c \: is \: downwars \: because \: its \: easy \: to \: pull \: the \: bike \: downwards \: but \: its \: hard \: to \: pull \: the \: bike \: upwards \: it \: would \: take \: more \: energy \: \gamma \\ \\$

Explanation:

<h3> $\infty \infty \: hope \: it \: helps \: you \: \\ have \: a \: good \: a \: day \: .$ </h3>

Slide my answer to see the full answer .

7 0

3 years ago